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sveticcg [70]
1 year ago
8

Determine the half-life of a nuclide that loses 38.0% of its mass in 407 hours. 590 hours 281 hours 291 hours 568 hour 204 hours

Physics
1 answer:
alexgriva [62]1 year ago
3 0

The half-life of a nuclide is 737 hrs

Let the initial mass of nuclide is N° = 100

The final mass of nuclide after 407 hours

= 100 - 38 = 62

The expression for decay constant is

λ = 1/t ln(N°/Nt)

Substitute the given values and calculate the decay constant as,

λ = 1/407hr ln(100/62)

λ = 9.4 x 10^-4/hr

Now, the half-life is calculated as,

t1/2 = 0.693/λ

= 0.693/9.4 x 10^-4/hr

t1/2 = 737 hrs

Hence, the half-life of a nuclide is 737 hrs

Learn more about Half-Life here:

brainly.com/question/25750315

#SPJ4

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Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of Na+ ions
iren [92.7K]

Answer:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

Explanation:

For this case we have the following info given:

Number of Na+ ions 5.7 x10^{11} ions

Each ion have a charge of +e and the crage of the electron is 1.6 x10^{-19}C

The time is given t = 7 ms if we convert this into seconds we got:

t = 7ms * \frac{1s}{1000 ms}= 0.007s

Now we can use the following formula given from the current passing thourhg a meter of nerve axon given by:

Q = Ne

Where N represent the number of ions, e the charge of the electron and Q the total charge

If we replace on this case we have this:

Q= 5.7x10^{11} * (1.6 x10^{-19}C) = 9.12x10^{-8} C

And from the general definition of current we know that:

I =\frac{Q}{t}

And since we know the total charge Q and the time we can replace:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

The current during the inflow charge in the meter axon for this case is 13.02 \mu A

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3 years ago
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3 years ago
The lightest car in the world was built in London and had a mass of less than 10 kg. it's maximum speed was 25.0 km/h. Suppose t
Softa [21]
25km/h = 6.94 m/s
suvat
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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23
Burka [1]

Answer:

a) v(2) = 3.9\,\frac{m}{s}, b) v(4) = -15.7\,\frac{m}{s}

Explanation:

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v(t) = 23.5 -9.8\cdot t

The velocities at given instants are, respectivelly:

v(2) = 3.9\,\frac{m}{s}

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Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to
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a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) ​K 2 +GmM( r 11− r 21)=6.9×10 7 J

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=K 2+U 2

⇒K 1− r 1GmM

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where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

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K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

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​=K 2 +GmM (r 11− r 21)=6.9×10 7 J

Learn more about Kinetic energy on:

brainly.com/question/12337396

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